Mom

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The Method of Moments Consider the inhomogeneous equation where is a linear operator, is known, and is to be determined. Let be expanded in a series of known functions , , ,...in the domain of , as where the are unknown constants to be determined. We shall call the expansion functions or basis functions. For exact solutions, the summation is infinite and the form a complete set of basis functions. For approximate solutions, finite summation is employed. Since is linear, we have . It is assume that a suitable inner product has been defined. Now definte a set of weighting functions, or testing functions, , , ,... in the range of , and take the inner product of the above equation with each . The result is , The set of equations can be written in matrix form as where , , If the matrix is nonsingular, its inverse exists. The are 1

Transcript of Mom

The Method of MomentsConsider the inhomogeneous equation

where is a linear operator, is known, and is to be determined. Let be expanded in a series of known functions , , ,...in the domain

of , as

where the are unknown constants to be determined. We shall call the expansion functions or basis functions. For exact solutions, the

summation is infinite and the form a complete set of basis functions.For approximate solutions, finite summation is employed. Since islinear, we have

.

It is assume that a suitable inner product has been defined. Nowdefinte a set of weighting functions, or testing functions, , , ,...in the range of , and take the inner product of the above equation witheach . The result is

,

The set of equations can be written in matrix form as

where

, ,

If the matrix is nonsingular, its inverse exists. The are

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then given by.

The solution may be exact or approximate, depending upon the choiceof and . The particular choice is known as Galerkin’smethod.

Example 1.

Exact solution :

Let Choose

and .Then,

We have

.

Then, .

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For ,

.

For ,

.

For ,

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,

which is the exact solution.

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Example 2:Pulse function:

Triangle function:

Let , , then

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HW#3: follow example 1, solve

by letting

Plot the result with the exact solution .

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Antenna ProblemsIn free space,

Therefore, there exist a dyadic satisfies

is called Green’s function. In general, in any structure, Green’sfunctions exist for computing the electric fields generated by a givensource.

Procedures:1. Specify excitation .2. Identify PEC.3. Identify the induced surface

current on the PEC due to

and let

4. Satisfy B. C. On the PEC.

5.

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where

For , we have

Similarly for slots on PEC, complementary formulation for magneticcurrents exist. Let

.

If two regions are separated by PEC with slots on it, then the slots canbe closed by putting equivalent magnetic currents on them. Let theregions be named and . Let be the normal vector pointing from to . Let the electric field on the slots be . Let the magnetic current in

region and be and respectively. Then,

At the slots, the magnetic field must be the same calculating from region and . We have

Let

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.

Then,

where

For , we have

Example: Slot-coupled microstripline-fed patch antennas

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Pocklington’s Integral Equation for a dipole

Let the length and the radius of the dipole be and , where suchthat we can ignore the circumferential current, therefore and

. Assume the conductivity of the dipole is vary large that is asurface current exist on the surface of the dipole. Since , we canalso assume doesn’t vary circumferentially. That is is only afunction of . Then

where

The approximation in the above equation is under the assumption thatthe surface current can be approximated by a line source at the centersince .

If only consider the field at surface of the dipole, .If an incident wave , the field will induce a current distribution onthe dipole surface. The induce current will produce scattered field .The boundary condition requires that on the surface of the dipole

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Integral Equations

The equation derived in previous section is called Integral Equation.The right hand side is the known incident field , while the left handside is the unknown current distribution to be solved.

Approximate by a series of known expansion functions such that

where are coefficients of the expansion functions and are unknown.For instance, choose impulse functions as defined below

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Point Matching

For conveniences, rewrite the integral equation as

Let be the mid-point of each segment, then

Rewriting in matrix form, we have

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where

For , we have

Source Modeling

1. Delta gap: or for .

2. Frill Generator:

where

3. Plane wave:

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The input impedance becomes if the applied voltage is

located at -th section.

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Weighted Residuals and the Method of Moments

Let weighting functions be a set of known functions.Instead of point-matching, the integration equation is integrated with theweighting function as follow:

The matrix formulation will be the same except

.

Obviously, if , the result is the same as point-matching.This general formulation is called the Method of Moments. If theweighting functions are chosen to be the same as the expansionfunction, i.e., , it’s called Galerkin’s method. In mostcases, Galerkin’s method gives better result than point-matching.

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Other Expansion and Weighting Functions

1. Triangle functions (piecewise linear)

2. Piecewise sinusoidal:

3. Sinusoidal interpolation:

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Adding Lumped Loads

If an ideal lumped load is inserted into awire antenna at coordinate havingcurrent , the electric field across canbe considered as a delta function havingamplitude . The original integralequation becomes

Therefore, the new impedance matrix is related to the unloaded oneby

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Method of Moment in Electrostatic Applications

Example: Stripline Capacitance

For a line charge , the voltage generated by this charge is

The potential produced by a distribution of charge is

If the distribution of the charge is discretized to sections as in the figure,let the charge density be constant in a cell and be denoted as . Let and be the coordinate and potential at the center of section . Let be the length of section . We have

where

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As shown in the figure, let the plates be divided to 36 sections with thesame section length . Eight sections are on the central strip and 28sections are on the top and bottom strips. Also let and .Therefore,

By solving the above equation, the capacitance can be found bydividing the total charge with voltage .

HW# 4 Let , find the capacitance. Compared with the formulain Pozar’s “Microwave Engineering”, pp 154–157.

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